The usual approach: let $q_n$ denote the probability of absorption starting from $n\geqslant0$, then $q_0=1$ and
$$q_n=pq_{n-1}+(1-p)q_{n+a}$$
for every $n\geqslant1$. Furthermore, since the only negative steps are $-1$ steps, to hit $0$ starting from $n$, one must hit $n-1$ starting from $n$, then hit $n-2$ starting from $n-1$, and so on until $0$. Thus, $q_n=(q_1)^n$ for every $n\geqslant0$. Can you deduce the value of $q_1$?
Likewise, assume that $q_1=1$ and let $t_n$ denote the mean absorption time starting from $n\geqslant0$ (if $q_1\ne1$, the mean absorption time is infinite), then $t_0=0$ and
$$t_n=1+pt_{n-1}+(1-p)t_{n+a}$$
for every $n\geqslant1$. Furthermore, since the only negative steps are $-1$ steps, the time to hit $0$ starting from $n$ is the sum of the time to hit $n-1$ starting from $n$, plus the time to hit $n-2$ starting from $n-1$, and so on until $0$. Thus, $t_n=nt_1$ for every $n\geqslant0$. Can you deduce the value of $t_1$?
Here is a partial answer, which proves the claim for Bernoulli random walks.
Let $S_i\in\{-1,1\}$ be independent step variables with $P(S_i=-1) =p$ and $P(S_i=1) = 1-p$, where $p<\frac12$ and denote $X_n = 1+\sum_{i=1}^nS_i$ as the random walk.
Furter, we introduce $T_0 = \inf\{n:X_n=0\}$ as the random variable that denotes the time the random walk first hits $0$. Then, by the Hitting Time Theorem,
$$ P(T_0=n) = \frac{P(X_n=0)}n $$
and consequently, the probability that the random walk hits $0$ at some point is given by
$$ P(T_0 < \infty) =\sum_{n=1}^\infty P(T_0=n)= \sum_{n=1}^\infty \frac{P(X_n=0)}n. $$
Since for even time instants, $P(X_{2k}=0)=0$ and odd time instants, $P(X_{2k+1}=0) = p^{k+1}(1-p)^k\binom{2k+1}{k+1}$, we get
$$ P(T_0 < \infty) = \sum_{k=0}^\infty \frac{p^{k+1}(1-p)^k\binom{2k+1}{k+1}}{2k+1} = \frac{p}{1-p},$$
where the last equality has been found by WolframAlpha. This means that the probability that the random walk never hits $0$ is given by $P(T_0 = \infty) = \frac{1-2p}{1-p}>0$.
This approach directly generalizes to the more general case, where $P(S = -1)+P(S=0)<\frac12$, which is always "dominated" by some Bernoulli random walk.
Best Answer
Let $a$ be the probability that you ever reach $n+1$ from $n$. Then we know that $$a=\frac23a^2 + \frac13^*$$ which gives us $a=\frac12$, as $a\neq 1$.
*How did I get this? Notice that the probability you ever reach $n+1$ in the next move is $\frac13$ -- and if you don't reach it in one move, then you go down a step. This means the probability of reaching it is $a^2$ because you need to move up twice, for a total of, well, what it says there.
So your answer is just $a^n=\frac1{2^n}$.